/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 In Problems 7鈥�12, show that ea... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 9

In Problems 7鈥�12, show that each series converges absolutely. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{2^{n}} $$

Short Answer

Expert verified
The series converges absolutely.

Step by step solution

01

Identify the Series

The given series is \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{2^{n}} \). This is an alternating series because it includes \((-1)^{n+1}\), which alternates the sign of each term.
02

Clarify Absolute Convergence

A series \( \sum a_n \) converges absolutely if the series \( \sum |a_n| \) converges. Here, \( a_n = (-1)^{n+1} \frac{n}{2^{n}} \), so \( |a_n| = \frac{n}{2^{n}} \). We need to determine if \( \sum_{n=1}^{\infty} \frac{n}{2^{n}} \) converges.
03

Apply the Ratio Test

To test absolute convergence, apply the Ratio Test: if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \), the series converges. Compute:\[\frac{a_{n+1}}{a_n} = \frac{n+1}{2^{n+1}} \cdot \frac{2^n}{n} = \frac{n+1}{2n}\]
04

Evaluate the Limit

Calculate the limit of \( \frac{n+1}{2n} \) as \( n \to \infty \):\[\lim_{n \to \infty} \frac{n+1}{2n} = \lim_{n \to \infty} \left( \frac{n}{2n} + \frac{1}{2n} \right) = \frac{1}{2} + 0 = \frac{1}{2}\]Since \( \frac{1}{2} < 1 \), the series \( \sum \frac{n}{2^{n}} \) converges.
05

Conclusion of Absolute Convergence

Since the series of absolute values \( \sum \frac{n}{2^{n}} \) converges by the Ratio Test, the original series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{2^{n}} \) converges absolutely.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alternating Series
An alternating series is a series whose terms alternate in sign. This means, for every term, the sign changes between positive and negative. Let's look at an example of an alternating series:
  • The series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{2^n} \) is alternating because it has the term \((-1)^{n+1}\), which switches the sign for every \(n\).
  • Alternating series can still converge even if individual terms keep switching from positive to negative.
One common method to determine convergence of alternating series is the Alternating Series Test. However, in the case of absolute convergence, we apply different tests like the Ratio Test, as we'll discuss in a later section.
Ratio Test
The Ratio Test is a powerful tool used to determine the convergence or divergence of a series. It evaluates the ratio of successive terms in the series and examines its limit as \(n\) approaches infinity. Here鈥檚 how it works:
  • For a series \(\sum a_n\), consider the ratio \(\frac{a_{n+1}}{a_n}\).
  • If \(\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1\), the series converges absolutely.
  • If the limit is greater than 1, the series diverges, and if it equals 1, the test is inconclusive.
In our specific example, the given series is \(\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{2^n}\). To apply the Ratio Test, we first examine the absolute values, thus considering \(\sum_{n=1}^{\infty} \frac{n}{2^n}\). By determining the limit \(\lim_{n \to \infty} \frac{n+1}{2n} = \frac{1}{2}\), we find it is less than 1, confirming absolute convergence.
Convergent Series
A convergent series is one where the sum of its terms approaches a finite number as more terms are added. Several methods can determine series convergence, each with specific requirements:
  • For a series to be absolutely convergent, the series of absolute values must converge. That is, \(\sum |a_n|\) must approach a finite limit.
  • The given series \(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{2^n}\) is proven to converge absolutely by showing \(\sum_{n=1}^{\infty} \frac{n}{2^n}\) itself converges using the Ratio Test.
Absolute convergence is particularly strong because it implies regular convergence, providing a stable foundation for evaluating series behavior. By proving absolute convergence, we gain confidence in the series鈥� behavior and the reliability of the sum鈥檚 finite limit.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find $$ \lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\sin \frac{k}{n}\right) \frac{1}{n} $$ Hint: Write an equivalent definite integral.

By writing \(1 / x=1 /[1-(1-x)]\) and using the known expansion of \(1 /(1-x)\), find the Taylor series for \(1 / x\) in powers of \(x-1\).

Find \(\lim _{n \rightarrow \infty} u_{n}\) of Problem 37 algebraically. Hint: Let \(u=\lim _{n \rightarrow \infty} u_{n}\). Then, since \(u_{n+1}=\sqrt{3+u_{n}}, u=\sqrt{3+u}\). Now square both sides and solve for \(u\).

In Problems 9-28, find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ 1+\frac{x+1}{2}+\frac{(x+1)^{2}}{2^{2}}+\frac{(x+1)^{3}}{2^{3}}+\cdots $$

In Problems 1-20, an explicit formula for \(a_{n}\) is given. Write the first five terms of \(\left\\{a_{n}\right\\}\), determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\) \(a_{n}=(2 n)^{1 / 2 n}\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.